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MATHEMATICS 29
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ENGLISH LANGUAGE 29
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Lecture3.1
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INTEGRATED SCIENCE 29
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SOCIAL STUDIES 29
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Lecture5.1
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Bece Past Questions & Answers – 2014 (Maths)
2014 BECE Mathematics (Maths) Past Questions – Paper One
1. If set N is a subset of set M, then
A. sets M and N have the same number of elements
B. some members of set N can be found in set M
C. no member of set N is in set M
D. all members of set N are in set M
D. all members of set N are in set M
The Venn diagram shows the number of pupils who offer Mathematics (M) and / or English in a class.
Use this information to answer Questions 2 and 3.
2. How many pupils offer Mathematics?
A. 10
B. 18
C. 25
D. 28
C. 25
3. How many pupils offer only one subject?
A. 3
B. 7
C. 18
D. 21
D. 21
4. Simplify: 12 – 7 – (– 5)
A. – 10
B. – 2
C. 0
D. 10
D. 10
5. Express 72 as a product of its prime factors
6. Find the smallest number which is divisible by 16 and 20?
A. 40
B. 80
C. 120
D. 160
B. 80
7. Convert to a base ten numeral.
A. 40
B. 43
C. 45
D. 73
D. 73
8. A pineapple which was bought for GH¢ 1.00 was sold at GH¢ 1.30. Calculate the profit percent.
A. 10%
B. 20%
C. 23%
D. 30%
D. 30%
9. Simplify
10. Two bells P and Q ring at intervals of 3 hours and 4 hours, respectively. After how many hours will the two bells first ring simultaneously (at the same time)?
A. 6 hours
B. 8 hours
C. 12 hours
D. 24 hours
C. 12 hours
11. A boy scores in a French test. Express his score as a percentage.
A. 17%
B 34%
C 68%
D 85%
C 68%
12. Arrange the following fractions in ascending order of magnitude
13. Kofi paid rent of GH¢ 1,800.00 each year. If the rent is 0.3 of his annual income, find his annual income.
A. GH¢ 600.00
B. GH¢5,400.00
C. GH¢ 6,000.00
D. GH¢ 18,000.00
C. GH¢ 6,000.00
14. I gave a storekeeper a GH¢10.00 note for goods I bought. He asked me for another 15Gp for ease of change. If he then gave me 50 Gp, how much did I pay for the goods?
A. GH¢ 9.35
B. GH¢ 9.45
C. GH¢ 9.65
D. GH¢ 10.65
C. GH¢ 9.65
15. Kojo can buy 15 shirts at GH¢ 4.00 each. If the price is increased to GH¢ 5.00, how many shirts can he now buy?
A. 12
B. 15
C. 19
D. 20
A. 12
16. A hall which is 8m long is represented on a diagram as 4 cm long. What is the scale of the diagram?
A. 1:200
B. 1:250
C. 1:400
D. 1:800
A. 1:200
17. Jane arrived at work at 7:55 am and left at 4:15 pm. For how long was she at work?
A. 7 hr 20 min
B. 7 hr 45 min
C. 8 hr 20 min
D. 8 hr 40 min
C. 8 hr 20 min
18. Given that (3.14 × 18) × 17.5 = 3.14 × (3p × 17.5), find the value of p
A. 3.0
B. 5.8
C. 6.0
D. 9.0
C. 6.0
The pie chart shows how Kwaku spends his monthly salary.
Use this information to answer Questions 19 to 21
19. Find the value of x
A. 65°
B. 75°
C. 85°
D. 100°
A. 65°
20. Kwaku earns GH¢ 630.00 a month. How much of this does he spend on food?
A. GH¢ 140.00
B. GH¢ 157.00
C. GH¢ 210.00
D. GH¢ 350.00
C. GH¢ 210.00
21. What percentage of his salary does he spend on rent and utilities?
A. 12.1%
B. 12.5%
C. 22.2%
D. 33.3%
C. 22.2%
22. In an enlargement with scale factor 2, which of the following statements is not true?
A. Each length is multiplied by 2
B. Each angle remains the same
C. The shape of the figure does not change.
D. The size of the figure does not change.
D. The size of the figure does not change.
23. Kofi, Kojo and Ama shared GH¢ 480,000.00 in the ratio 3:5:4. How much did Ama receive?
A. GH¢ 160,000.00
B. GH¢ 200,000.00
C. GH¢ 218,181.81
D. GH¢ 342,859.14
A. GH¢ 160,000.00
24. If w = 12, x = 5, y = 6 and z = 4, find the value of wx – yz.
A. 18
B. 27
C. 36
D. 84
C. 36
25. A man was 24 years old when his son was born. Now he is three times as old as his son. Find the age of the son.
A. 6 years
B. 12 years
C. 18 years
D. 36 years
B. 12 years
26. There are 20 identical balls in a box. Twelve are blue and the rest are green. If one ball is taken at random from the box, find the probability that the ball is green.
27. Using the following mapping, find the missing numbers p and q.
x 1 2 3 4 5 6
↓ ↓ ↓ ↓ ↓ ↓ ↓
y 3 5 p 9 11 q
A. p = 6, q = 12
B. p = 6, q = 13
C. p = 7, q = 12
D. p = 7, q = 13
D. p = 7, q = 13
28. The perimeter of a rectangle is 24 cm. If the length is 7 cm, find its width.
A. 3 cm
B. 5 cm
C. 10 cm
D. 12 cm
B. 5 cm
29. A boy walks on a bearing 070°. Which of the following diagrams show his direction?
B.
30. How many faces has a cube?
A. 4
B. 6
C. 8
D. 12
B. 6
31. The diameter of a circular tray is 28 cm. Find the area of the tray. [Take π = 22/7]
A 44 cm2
B 88 cm2
C 154 cm2
D 616 cm2
D 616 cm2
32. Calculate the volume of a cylinder with radius 7 cm and height 10 cm. [Take π = 22/7]
A. 220 cm3
B. 440 cm3
C. 1,540 cm3
D. 3,080 cm3
C. 1,540 cm3
Use the diagram below to answer questions 33 and 34
33. Find the value of e.
A. 38°
B. 40°
C. 88°
D. 92°
D. 92°
34. Find the angle marked d
A. 38°
B. 40°
C. 48°
D. 88°
B. 40°
35. A 3.6 m long string is to be cut into pieces, each of length 40 cm. How many pieces can be cut from the string?
A. 4
B. 6
C. 8
D. 9
D. 9
36. Solve the inequality
A. x ≤ 10
B. x ≥ 10
C. x ≤ 40
D. x ≥ 40
A. x ≤ 10
37. The point P (5, 4) is reflected in the y-axis. Find its image.
A. (– 5, 4)
B. (5, –4)
C. (–4, 5)
D. (4, –5)
A. (– 5, 4)
38. If , find the value of x.
A –1
B 1
C 7
D 12
C 7
39. Find the gradient of the line which passes through the points M(–1, 2) and N(6, –3)
A (-5)/7
40. Find the next two terms in the sequence 11, 7, 3, –1, ¬___, ___.
A. 5, 9
B. 3, 7
C. –4, –9
D. –5, –9
D. –5, –9
2014 BECE Mathematics (Maths) Past Questions – Paper Two
1. (a) P = {factors of 30}
Q = {Multiples of 5 less than 40}
Find P ∩ Q
(b) A trader saved GH¢ 200.00 for 3 years at 12% simple interest per annum.
What will be the total amount in the trader’s account at the end of the 3 years?
(c) Evaluate and leave your answer in standard form.
1. (a) P = {1, 2, 3, 5, 6, 10, 15, 30}
Q = {5, 10, 15, 20, 25, 30, 35}
P∩Q = {5, 10, 15, 30}
1. (b)
Total amount = Interest + Principal
= GHC 72.00 + GHC 200.00
= GHC 272.00
1. (c)
1st Method
1 (c) 2nd Method
1 (c) 3rd Method
2. (a) (i) Ama scored 82, 74 and 90 in three tests. What mark should she score in the fourth test so that her average mark for the four tests would be 85?
(ii) What was her median score in the four tests?
(b)
In the diagram (AD) ̅ is parallel to (EG) ̅, angle CFG = 40° and triangle BCF is isosceles.
Find the value of :
(i) angle CBF
(ii) angle DCF;
(iii) x
2. (a) Let x = Ama’s score in the fourth test
Method 1
2. (a) (i) Method 2
Total marks = No. of marks × mean mark
= 4 × 85
= 340
Sum of first 3 marks = 82 + 74 + 90
= 246
Ama’s fourth mark = Total mark – sum of first three
= 340 – 246
= 94
(a) (ii) Median score
Scores arranged in order gives 74, 82, 90, 94
(b) (i) Since angles BCF and CFG are alternate angles,
⇒ Angle BCF = 40°
Now, since base angles of isosceles triangle BFC are equal,
⇒ Angle CBF = 40°
(ii) angle DCF + angle BCF = 180° (angles at a point on a straight line = 180°)
⇒ angle DCF + 40° = 180°
⇒ angle DCF = 180° – 40°
= 140°
(iii) 2x + 40° + 40° = 180° (Sum of interior angles of a triangle = 180°)
2x + 80° = 180°
2x = 180° – 80°
2x = 100°
2x/2 = 100/2
x = 50
3. (a) Solve for x, if
(b) The following shows the distribution of marks of students in an examination.
6 43 26 18 27
42 8 22 31 39
55 44 37 47 59
10 12 36 53 48
(i) Make a stem-and-leaf plot of the marks above
(ii) Find the probability of selecting a student who scored between 40 and 50.
(iii) Find the number of students who passed the examination, if the pass mark was 30.
3. (a) Solve for x,
3.(b) (i) Stem-and-leaf plot
Stem | Leaf |
0 | 6, 8 |
1 | 0, 2, 8 |
2 | 2, 6, 7 |
3 | 1, 6, 7, 9 |
4 | 2, 3, 4, 7, 8 |
5 | 3, 5, 9 |
(ii) Probability of selecting a student who scored between 40 and 50
(iii) Number of students who passed, if the pass mark was 30
= n (31, 36, 37, 39, 42, 43, 44, 47, 48, 53, 55, 59)
= 12 students
4. (a) A box has length 8.0 cm, width 5.0 cm and height 10.0 cm.
Find the:
(i) total surface area of the box
(ii) the volume of the box.
(b) (i) Using a scale of 2cm to 1 unit on both axes, draw two perpendicular axes Ox and Oy on a graph sheet.
(ii) On the same graph sheet mark the x-axis from –5 to 5 and the y-axis from – 6 to 6
(iii) Plot and join the points A(0, 3), B(2, 3), C(4, 5) to form triangle ABC.
(iv) Draw the image A1B1C1 of triangle ABC under a translation by the vector -1-1
(v) Draw the image A2B2C2 of triangle ABC under a reflection in the x – axis
4. (a) (i) Let length = l, width = w, height = h
Total surface area = 2lw + 2lh + 2wh,
= (2× 8cm × 5cm) + (2 × 8cm × 10cm) + (2 × 5cm × 10cm)
= 80cm2 + 160cm2 + 100cm2
= 340 cm2
(ii) Volume = l × w × h
= 8cm × 5cm × 10cm
= 400 cm3
4 (b)
5. (a) Using a ruler and a pair of compass only;
(b) (i) Measure the radius of the circle.
(ii) Calculate the circumference of the circle, correct to 3 significant figures.
[Take π = 3.14]
5. (a)
(b) (i) Radius = 4.0cm (or 4.1cm)
(ii) If r = 4.0 cm
C = 2 π r
= 2 × 3.14 × 4 cm
= 25.12 cm
Or if r = 4.1cm
C = 2 × 3.14 × 4.1 cm
= 25.748 cm
6. Factorize completely 6xy – 3y + 4x – 2
(b)
The diagram shows a ladder AB which leans against a vertical wall PQ at B.
If |PB| is 8 m, and the other end of the ladder is 6 m away from the foot of the wall (at P), find the length of the ladder (|AB|)
(c) Kojo had 1800 bags of rice in stock for sale. In January, he sold ⅔ of it.
In February, he sold ¾ of what was left.
(i) What fraction of the stock of rice did he sell
(α) in February?
(β) in January and February?
(ii) How many bags of rice were left unsold, by the end of February?
6. (a) 6xy – 3y + 4x – 2
3y(2x – 1) + 2(2x – 1)
(2x – 1) (3y + 2)
(b) The length of the ladder AB forms the hypotenuse of the right-angled triangle ABP
From the Pythagorean theorem,
|AB|2 = |AP|2 + |BP|2
= (6)2 + (8)2
= 36 + 64
|AB|2 = 100
⇒ |AB| =
= 10 m
The length of the ladder AB is 10 m
6. (c) Method 1
No. of bags sold in January =
= 2 × 600
= 1200
No. of bags left = 1800 – 1200
= 600
No. of bags sold in February =
= 3 × 150
= 450
(i) (?) Fraction of bags sold in February =
=
=
(i) (?) Fraction of bags sold in Jan and Feb =
=
=
(ii) No.of bags left unsold by the end of February = 1800 – 1650
= 150
6. (c) Method 2
Fraction sold in January =
Fraction left = 1 –
= –
=
=
(i) (?) Fraction sold in February = of fraction left
= x
= x
Fraction sold in Feb. =
(i) (?) Fraction sold In January and February
(ii) No. of bags left unsold by end of February
= Fraction left unsold × Total no. of bags
But fraction left unsold =
Therefore No. of bags left unsold by end of February
=
= 1 × 150 bags
= 150 bags